Answer:
[tex](x+3)^2+(y-6)^2=9[/tex]
Step-by-step explanation:
The equation of a circle in standard form is written as
[tex](x-a)^2+(y-b)^2=r^2[/tex]
where
(a, b) are the coordinates of the center
r is the radius of the circle
In this problem, the circle is centered at (-3,6), so we have
a = -3
b = 6
Also, we know that the circle is tangent to the y-axis, and the x-coordinate of its center is 3: this means that the horizontal distance between the point of the circle tangent to the y-axis and the centre of the circle is 3, therefore the radius of the circle is 3.
So, the equation of this circle is:
[tex](x-(-3))^2+(y-6)^2=3^2[/tex]
[tex](x+3)^2+(y-6)^2=9[/tex]
